Question

A small object of mass m= 90 kg slides down a spherical dome of radius R=12 m without any friction. It starts off at the top (polar angle θ=0) at zero speed. Use g=10 m/s2. (See figure)

(a) What is the magnitude of the force (in Newtons) exerted by the dome on the mass when it is at the top, at θ=0∘?

N(θ=0∘)=


(b) What is the magnitude of the force (in Newton) exerted by the dome on the mass when it is at θ=30∘?

N(θ=30∘)=

(c) At what angle θ0 does the sliding mass take off from the dome? answer in degrees (0∘≤θ0≤90∘; )

θ0=

Answer 1

Gravity exerts a normal force on the dome of
forcenormal=mg*CosTheta where theta is measured from the vertical to the point of contact.

c. when mv^2/r>normal force, it takes off.

now v is the result of gravity PE converting to KE.
let Theta=90 deg=zero PE point, or referece.
then at the top, PE=mgr
PE at any other point then is
PE(Theta)=mgr*cosTheta
so KE= mgr(1-cosTheta)

or 1/2 m v^2=mgr (1-cosTheta)
v^2 = 2gr (1-cosTheta)

finally, when mv^2/r >mgcosTheta
2g ((1-cosTheta)> gcosTheta
1/2 -1/2 cosTheta>cosTheta
or -3/2 cosTheta>-1/2
or cosTheta<1/3 and at the angle
Theta=arccos 1/3= 70.58 degrees it flys off.

check my thinking and my work.

Answer 2

does not be 2g(1-costheta)>gcostheta
2-2c0stheta>costheta?
you wrote 1/2-1/2costheta>cosTheta

Answer 3

@bobpursley u r crct man.. but make the changes pointed by asha..! :) i got dem crct

Answer 4

I didn't get the 2nd part, If I plug in 50*10*cos(30) , i get the answer wrong.

Answer 5

for second part it is m*g*tan theta

Answer 6

@ks did you got mass pushed by spring?

Answer 7

@KS can you explain why it is m*g*tan(theta)?